Assuming 100% dissociation of K3PO4 you get each molecule of K3 PO4 dissociates into four particles: 3 K4 cations and 1 PO4 - anion. That means that Van't Hoff factor, i, is 4. i = 4. So, the decrease in freezing point of the water solution is Delta Tf = i * Kf * m. Where Kf is the molal cryoscopic constant of water = 1.86 °C /m; and the molality m = 2.59 m => Delta Tf = 4 * 1.86 °C / m * 2.59 m = 19.3 °C. And the freezing point is the normal freezing points less 19.3°C = 0°C - 19.3°C = - 19.3°C. The increase in the boiling point Delta Tb = i * kb * m = 4 * 0.512 °C /m * 2.59 m = 5.3°C => <span>Tb = normal boiling point + 5.3°C = 100°C + 5.3°C = 105.3°C.</span>
Answer:
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Explanation:
Answer:
An input of heat energy Yr don't you know stupied
Na3PO4 + 3 KOH = 3 NaOH + K3PO4
Answer:
the new concentration is 0.60M
Explanation:
The computation of the new concentration is shown below;
We know that
M1V1=M2V2
(3.0M) (10.0 mL) = M2 (50.0mL)
30 = M2 (50.0mL)
So, M2 = 0.60 M
Hence, the new concentration is 0.60M
The same is considered and relevant
